LIMITS OF FUNCTIONS USING THE SQUEEZE PRINCIPLE
The following problems involve the algebraic computation of limits using the
Squeeze Principle, which is given below.
SQUEEZE PRINCIPLE : Assume that functions f , g , and h satisfy
and
.
Then
.
(NOTE : The quantity A may be a finite number,
be a finite number,
, or
, or
. The quantitiy L may
.)
The Squeeze Principle is used on limit problems where the usual algebraic methods
(factoring, conjugation, algebraic manipulation, etc.) are not effective. However, it
requires that you be able to ``squeeze'' your problem in between two other ``simpler''
functions whose limits are easily computable and equal. The use of the Squeeze
Principle requires accurate analysis, deft algebra skills, and careful use of inequalities.
SOLUTIONS TO LIMITS USING THE SQUEEZE PRINCIPLE
SOLUTION 1 : First note that
because of the well-known properties of the sine function. Since we are computing the
limit as x goes to infinity, it is reasonable to assume that x > 0 . Thus,
.
Since
1
,
it follows from the Squeeze Principle that
.
SOLUTION 2 : First note that
because of the well-known properties of the cosine function. Now multiply by -1,
reversing the inequalities and getting
or
.
Next, add 2 to each component to get
.
Since we are computing the limit as x goes to infinity, it is reasonable to assume that x
+ 3 > 0. Thus,
.
Since
,
it follows from the Squeeze Principle that
2
.
SOLUTION 3 : First note that
because of the well-known properties of the cosine function, and therefore
.
Since we are computing the limit as x goes to infinity, it is reasonable to assume that 3
- 2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting
,
or
.
Since
,
it follows from the Squeeze Principle that
.
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SOLUTION 4 : Note that
DOES NOT EXIST since values of
oscillate between -1 and +1 as x approaches 0 from the left. However, this
does not exist ! ? #. Indeed, x3 < 0
does NOT necessarily mean that
and
for x < 0. Multiply each component by x3, reversing the inequalities and getting
or
.
Since
,
it follows from the Squeeze Principle that
.
SOLUTION 5 : First note that
,
so that
4
and
.
Since we are computing the limit as x goes to infinity, it is reasonable to assume that
x+100 > 0. Thus, dividing by x+100 and multiplying by x2, we get
and
.
Then
=
=
=
=
.
Similarly,
=
.
Thus, it follows from the Squeeze Principle that
=
5
(does not exist).
SOLUTION 6 : First note that
,
so that
,
,
and
.
Then
=
=
=
=5.
Similarly,
6
=5.
Thus, it follows from the Squeeze Principle that
=5.
SOLUTION 7 : First note that
and
,
so that
and
.
Since we are computing the limit as x goes to negative infinity, it is reasonable to
assume that x-3 < 0. Thus, dividing by x-3, we get
or
.
Now divide by x2 + 1 and multiply by x2 , getting
7
.
Then
=
=
=
=
=0.
Similarly,
=0.
It follows from the Squeeze Principle that
=0.
SOLUTION 8 : Since
8
=
and
=
,
it follows from the Squeeze Principle that
,
that is,
.
Thus,
.
SOLUTION 9 : a.) First note that (See diagram below.)
area of triangle OAD < area of sector OAC < area of triangle OBC .
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The area of triangle OAD is
(base) (height)
.
The area of sector OAC is
(area of circle)
.
The area of triangle OBC is
(base) (height)
.
It follows that
or
10
.
b.) If
, then
and
, so that dividing by
.
Taking reciprocals of these positive quantities gives
or
.
Since
,
it follows from the Squeeze Principle that
.
SOLUTION 10 : Recall that function f is continuous at x=0 if
i.) f(0) is defined ,
ii.)
exists ,
and
iii.)
.
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results in
First note that it is given that
i.) f(0) = 0 .
Use the Squeeze Principle to compute
. For
we know that
,
so that
.
Since
it follows from the Squeeze Principle that
ii.)
.
Finally,
iii.)
,
confirming that function f is continuous at x=0 .
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